The matrix is not properly formatted.
However, I'm able to rearrange the question as:
![\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C-2%263%265%7C3%5C%5C3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
Operations:
Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.
Answer:
![\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5C%5C0%265%267%7C1%5C%5C0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The first operation:
This means that the new second row (R2) is derived by:
Multiplying the first row (R1) by 2; add this to the second row
The row 1 elements are:
![\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D)
Multiply by 2
![2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]](https://tex.z-dn.net/?f=2%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%262%7C-2%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%265%7C3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&5&7|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%265%267%7C1%5Cend%7Barray%7D%5Cright%5D)
The second operation:
This means that the new third row (R3) is derived by:
Multiplying the first row (R1) by -3; add this to the third row
The row 1 elements are:
Multiply by -3
![-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]](https://tex.z-dn.net/?f=-3%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%7C-1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D)
Add to row 2 elements are: ![\left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%26-3%26-3%7C3%5Cend%7Barray%7D%5Cright%5D%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%264%7C1%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%261%7C4%5Cend%7Barray%7D%5Cright%5D)
Hence, the new matrix is:
Answer:
-79.8
Step-by-step explanation:
The change can be found by subracting the ending distance from the beginning distance.
In this case, its 200.6-120.8= 79.8
It is negative because the distance in the end became lower than the beginning distance.
Hope this helps!
The answer is the second option which is : - 29
If you have supplementary angles, that means that the angles have to add up to 180°. So x+(x+8)=180° So I would combine like terms 2x+8=180 Now subtract by 8 on both sides which gives me 2x=172 now divide by 2 and x equals 86°! I hoped this helped you!
Answer:
D
Step-by-step explanation:
because if you unfold the left corner at the top it just becomes a regular rectangle, 15 by 10. 15 x 2 plus 10 x 2 = 54. and 54 x 1.19 is $64.26 therefor it is D.
I actually did the math this time i promise.
